Question

假设我有一个x表示重复测量的数组（1-4），y表示处理（A，B），z表示时间点（1-3）

x <- c(2,2,4,15,17,13,3,10,3,4,11,14,1,3,19,6,13,6,12,18,9,13,12,16)
dim(x) <- c(4,2,3)

, , 1

     [,1] [,2]
[1,]    2   17
[2,]    2   13
[3,]    4    3
[4,]   15   10

, , 2

     [,1] [,2]
[1,]    3    1
[2,]    4    3
[3,]   11   19
[4,]   14    6

, , 3

     [,1] [,2]
[1,]   13    9
[2,]    6   13
[3,]   12   12
[4,]   18   16

我想创建一个新数组，其每次复制的次数都大于该治疗和时间点组合的所有其他重复次数：

, , 1

     [,1] [,2]
[1,]    2    0 #both 4 and 15 are bigger then 2, so for 1,1,1 the result is 2
[2,]    2    1
[3,]    1    3 #15 is the only replicate bigger than 4 so result for 3,1,1 is 1
[4,]    0    2

, , 2

     [,1] [,2]
[1,]    3    3
[2,]    2    2
[3,]    1    0
[4,]    0    1

, , 3

     [,1] [,2]
[1,]    1    3 
[2,]    3    1 
[3,]    2    2 
[4,]    0    0

Answer 1

apply可以做到这一点，在每一栏（2）和阶层（3）内行动：

## recreate your data array:
arr <- c(2,2,4,15,17,13,3,10,3,4,11,14,1,3,19,6,13,6,12,18,9,13,12,16)
dim(arr) <- c(4,2,3)

## one liner using apply
apply(arr, 2:3, function(x) sapply(x, function(y) sum(y < x) ) )

#, , 1
#
#     [,1] [,2]
#[1,]    2    0
#[2,]    2    1
#[3,]    1    3
#[4,]    0    2
#
#, , 2
# 
#     [,1] [,2]
#[1,]    3    3
#[2,]    2    2
#[3,]    1    0
#[4,]    0    1
# 
#, , 3
# 
#     [,1] [,2]
#[1,]    1    3
#[2,]    3    1
#[3,]    2    2
#[4,]    0    0

Answer 2

这里你去......如果你的问题是错误的措辞（我上面怀疑），那么你需要使用＆＃34;＆lt;＆＃34;而不是＆＃34;＆gt;＆＃34;。

a <- array(rnorm(24), dim= c(4,2,3))

cnts <- function(a) {
  a2 <- array(NA, dim= dim(a))
  for (i in 1:dim(a)[3]) {
    for (j in 1:dim(a)[2]) {
      for (k in 1:length(a[,j,i])) {
        a2[k,j,i] <- sum(a[k,j,i] > a[,j,i])
      }
    }
  }
  return(a2)
}

计算大于r中数组列中每个值的值的数量

2 个答案: